Find the values of p and q so that x4

Solution:

Here, $f(x)=x^{4}+p x^{3}+2 x^{2}-3 x+q$

$g(x)=x^{2}-1$

first, we need to find the factors of $x^{2}-1$

$\Rightarrow x^{2}-1=0$

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

$\Rightarrow(x+1)$ and $(x-1)$

From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0

Let us take, x + 1

⟹ x + 1 = 0

⟹ x = -1

Substitute the value of x in f(x)

$f(-1)=(-1)^{4}+p(-1)^{3}+2(-1)^{2}-3(-1)+q$

= 1 - p + 2 + 3 + q

= -p + q + 6 .... 1

Let us take, x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute the value of x in f(x)

$f(1)=(1)^{4}+p(1)^{3}+2(1)^{2}-3(1)+q$

= 1 + p + 2 - 3 + q

= p + q  ..... 2

Solve equations 1 and 2

- p + q = - 6

p + q = 0

2q = - 6

q = - 3

substitute q value in equation 2

p + q = 0

p - 3 = 0

p = 3

the values of are p = 3 and q = – 3